About $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ and $\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C}$

Question

We know $\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C} = \mathbb{C}$ and $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} = \mathbb{C}^2$. This contradicts my current understanding. I've spent time looking up answers, but can't seem to make the situation clear.

My understanding is the tensor products here are the best $\mathbb{C}$-modules to embed the $\mathbb{R}$-module $\mathbb{C}$ (or $\mathbb{C}$-module $\mathbb{C}$) into such that it's scalar are extended to $\mathbb{C}$. Specifically $\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C} $ and $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ both satisfy universal properties.

However why does $\mathbb{C}$ not satisfy the universal property for the $\mathbb{R}$-module $\mathbb{C}$ that makes it necessary to embed it in $\mathbb{C}^2$ instead?

Answer 1

The thing to keep in mind is that the universal property of the tensor product tells us that $\mathbb C\otimes_{\mathbb R} \mathbb C$ is the free $\mathbb C$-module on the $\mathbb R$-module $\mathbb C$. Free means here in particular that you don't want to introduce any relations that do not hold before extending scalars. That's why $\mathbb C$ as a module over itself is not free on the $\mathbb R$- module $\mathbb C$: the equation $\mathrm i^2+1=0$ shows that $\mathrm i$ and $1$ are linearly dependent over $\mathbb C$, but they are not linearly dependent over $\mathbb R$.

Answer 2

Well, in $\Bbb C\otimes_{\Bbb C}\Bbb C$, each element has the form $z\otimes z' = zz'\otimes 1$ and so can be viewed as element $zz'$ of $\Bbb C$.

In $\Bbb C\otimes_{\Bbb R}\Bbb C$, such an identification is not possible, so $z\otimes z'$ can be viewed as element $(z,z')$ of $\Bbb C^2$.